A vibration magnetometer placed in magnetic meridian has a small...

The time period T of oscillation of a magnet is given by (T=2 \pi \sqrt{\frac{I}{M B}}) where, I = Moment of inertia of the magnet about the axis of rotation $M=$ Magnetic moment of the magnet $B=$ Uniform magnetic field As I, B remains the same (\therefore \quad T \propto \frac{1}{\sqrt{B}}) or (\frac{T_2}{T_1}=\sqrt{\frac{B_1}{B_2}}) According to given problem, $$\begin{aligned} B_1 & =24 \mu \mathrm{T} \\ B_2 & =24 \mu \mathrm{T}-18 \mu \mathrm{T}=6 \mu \mathrm{T} \\ T_1 & =2 \mathrm{~s} \\ \therefore \quad T_2 & =(2 \mathrm{~s}) \sqrt{\frac{(24 \mu \mathrm{T})}{(6 \mu \mathrm{T})}}=4 \mathrm{~s} \end{aligned}$$)$